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Q.

The angular momentum of electron in ‘d’, orbital is equal to(where h = h/2π)

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a

0 h

b

$\sqrt{2} h$

c

$2 \sqrt{3} h$

d

$\sqrt{6} h$

answer is C.

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Detailed Solution

$ℓ = \sqrt{ℓ (ℓ + 1)} \cdot \frac{h}{2 π}$
in d – orbital l=2
$l = \sqrt{2 (2 + 1)} . \frac{h}{2 π} \Rightarrow \sqrt{6} . \frac{h}{2 π}$

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